Resolvent expansions for the Schrödinger operator on a graph with rays

Transcription

1 Resolvent expansions for the Schrödinger operator on a graph with rays Kenichi ITO (Kobe University) joint work with Arne JENSEN (Aalborg University) 6 September 2017

2 Purpose of this talk: Resolvent expansion Let (G, E G ) be a graph with a finite number of infinite rays, and we consider the Schrödinger operator where, for any u: G C, ( G u)[x] = H = G + V, {x,y} E G ( u[x] u[y] ) for x G. The purpose of this talk is to compute the expansion ( H + κ 2 ) 1 κ 2 G 2 + κ 1 G 1 + G 0 + κg 1 + as κ 0. It is well known by Jensen Kato (1979) that this provides e ith N e itλ jp λj + t 1/2 C 1 + t 3/2 C 0 + j=1 as t ±. 1

3 Graph with rays Let (K, E K ) be a connected, finite, undirected and simple graph, and let (L α, E α ), α = 1,..., N, be N copies of the discrete half-line: L α = N = { 1, 2,... }, E α = { {n, n + 1}; n L α }. Fix any x α K for α = 1,..., N, and define the graph (G, E G ) by G = K L 1 L N, E G = E K E 1... E N { {x1, 1 (1)},..., { x N, 1 (N)}}. Note that two different half-lines (L α, E α ) and (L β, E β ), α β, could be jointed to the same vertex x α = x β K. 2

4 Function spaces Our Hilbert space is H = l 2 (G) = [ l 2 (K) ] [ l 2 (L 1 ) ] [ l 2 (L N ) ]. We also consider for any s R L s = l 1,s (G) = [ l 1 (K) ] [ l 1,s (L 1 ) ] [ l 1,s (L N ) ], (L s ) = l, s (G) = [ l (K) ] [ l, s (L 1 ) ] [ l, s (L N ) ]. For example, if we define n (α) : G C, 1 (α) : G C as n (α) [x] = { m for x = m Lα, 0 for x G \ L α, 1 (α) [x] = { 1 for x Lα, 0 for x G \ L α, respectively, then n (α) (L 1 ) and 1 (α) (L 0 ). 3

5 Class of perturbations Note that any local potential V has a factorization V[n] = v[n]u[n]v[n]; v[n] := V[n], U[n] := sgn V[n]. Assumption (A). Let β 1, and K be an abstract Hilbert space. Assume that V B(H) is self-adjoint, and that there exist and such that an injective operator v B(K, L β ) a self-adjoint and unitary operator U B(K) V = vuv B ( (L β ), L β). 4

6 Examples. 1. Let V be a multiplication operator, and assume that Then V satisfies (A). V[n] C n 1 2β ɛ for some ɛ > Let V B(H) be self-adjoint, and assume that it extends to an operator in B ( l 2, β 1/2 ɛ (G), l 2,β+1/2+ɛ (G) ) for some ɛ > 0. Then V satisfies (A). 5

7 Generalized eigenspaces Under (A) we let H = H 0 + V, R(κ) = ( H + κ 2) 1. H is a bounded and self-adjoint on H, and σ ess (H) = [0, 4]. Define the generalized zero eigenspace as Ẽ = { Ψ (L β ) ; HΨ = 0 }. We can actually show that Ẽ Cn (1) + + Cn (N) + C1 (1) + + C1 (N) + L β 2. It would be natural to consider the subspaces E = Ẽ ( C1 (1) + + C1 (N) + L β 2), E = Ẽ Lβ 2. 6

8 Definition. The threshold z = κ 2 = 0 is said to be 1. a regular point, if E = E = {0}; 2. an exceptional point of the first kind, if E E = {0}; 3. an exceptional point of the second kind, if E = E {0}; 4. an exceptional point of the third kind, if E E {0}. Remark. We can actually show that dim(ẽ/e) + dim(e/e) = N, 0 dim E <. 7

9 Generalized eigenprojections Definition. We call a basis { Ψ γ }γ E a bound basis. It is said to be orthonormal, if for any γ and γ Ψ γ, Ψ γ = δ γγ. The orthogonal bound projection P is defined as P = γ Ψ γ Ψ γ. 8

10 Definition. We call { Ψ γ }γ E a resonant basis, if { [Ψ γ ] } γ E/E forms a basis. It is said to be orthonormal, if 1. for any γ and Ψ E one has Ψ, Ψ γ = 0; 2. there exits an orthonormal system { c (γ)} γ CN such that Ψ γ N α=1 c (γ) α 1 (α) L β 2. The orthogonal resonant projection P is defined as P = γ Ψ γ Ψ γ. 9

11 Definition. We call { Ψ γ }γ Ẽ a non-resonant basis, if { [Ψ γ ] } γ Ẽ/E forms a basis. It is said to be orthonormal, if 1. for any γ and Ψ E one has Ψ, Ψ γ = 0; 2. there exits an orthonormal system { c (γ)} γ CN such that Ψ γ N α=1 c (γ) α n (α) C1 (1) + + C1 (N) + L β 2. The orthogonal non-resonant projection P is defined as P = γ Ψ γ Ψ γ. 10

12 Main results We set B s = B ( L s, (L s ) ) = B ( l 1,s (G), l, s (G) ). Theorem 1. Assume that 0 is a regular point, and that (A) is fulfilled for some integer β 2. Then R(κ) = β 2 j=0 κ j G j + O(κ β 1 ) in B β 2 with G j B j+1 for j even, and G j B j for j odd. The coefficients G j can be computed explicitly. In particular, G 2 = P = 0, G 1 = P = 0. 11

13 Theorem 2. Assume that 0 is an exceptional point of the first kind, and that (A) is fulfilled for some integer β 3. Then R(κ) = β 4 j= 1 κ j G j + O(κ β 3 ) in B β 1 with G j B j+3 for j even, and G j B j+2 for j odd. The coefficients G j can be computed explicitly. In particular, G 2 = P = 0, G 1 = P = 0. 12

14 Theorem 3. Assume that 0 is an exceptional point of the second kind, and that (A) is fulfilled for some integer β 4. Then R(κ) = β 6 j= 2 κ j G j + O(κ β 5 ) in B β 2 with G j B j+3 for j even, and G j B j+2 for j odd. The coefficients G j can be computed explicitly. In particular, G 2 = P 0, G 1 = P = 0. 13

15 Theorem 4. Assume that 0 is an exceptional point of the third kind, and that (A) is fulfilled for some integer β 4. Then R(κ) = β 6 j= 2 κ j G j + O(κ β 5 ) in B β 2 with G j B j+3 for j even, and G j B j+2 for j odd. The coefficients G j can be computed explicitly. In particular, G 2 = P 0, G 1 = P = 0. 14

16 Theorems 1 4 justify the classification of threshold types only by the growth properties of the eigenfunctions: Corollary 5. The threshold type determines and is determined by the coefficients G 2 and G 1. Once an expansion is guaranteed, the general theory immediately implies: Corollary 6. The coefficients G j satisfy HG j = G j H = 0 for j = 2, 1, HG 0 = G 0 H = I P, HG j = G j H = G j 2 for j 1. 15

17 The free operator Let h K be the Dirichlet Laplacian on K: For any u: K C (h K u)[x] = {x,y} E K ( u[x] u[y] ) + N α=1 where s α [x] = 1 if x = x α and s α [x] = 0 otherwise. s α [x]u[x] for x K, Let h α be the Dirichlet Laplacian on L α : For any u: L α C { 2u[1] u[2] for n = 1, (h α u)[n] = 2u[n] u[n + 1] u[n 1] for n 2. Remark. The Dirichlet boundary conditions are set outside the graphs K and L α, respectively. 16

18 Then we define the free operator H 0 on G as a direct sum H 0 = h K h 1 h N. The operator H 0 does not coincide with G. In fact, G = H 0 + J, J = N α=1 ( ) s α f α + f α s α, where f α [x] = 1 if x = 1 (α) and f α [x] = 0 otherwise. However, H 0 is simpler and more useful than G, since it does not have zero eigenvalue or zero resonance. Since J satisfies (A), our setting covers the graph Laplacian G as a perturbation of H 0. 17

19 Expansion of the free resolvent We can directly compute an expansion of the free resolvent: Proposition 7. There exist 0 < λ 1 λ 2 λ #K such that σ ac (H 0 ) = [0, 4], σ sc (H 0 ) =, σ pp (H 0 ) = { λ 1,..., λ #K }. In addition, for any integer ν 0, the resolvent R 0 (κ) = ( H 0 +κ 2) 1 has an expansion: R 0 (κ) = ν κ j G 0,j + O(κ ν+1 ) in B ν+2 j=0 with G 0,j B j+1 for j even, and G 0,j B j for j odd. 18

20 The zeroth and first order coefficients We set M 0 = U + v G 0,0 v B(K). Let a C be the pseudo-inverse of a C defined as { a 0 if a = 0, = a 1 if a 0. We define the pseudo-inverse M 0 by the operational calculus. Theorem 8. In the regular case, one has G 0 = G 0,0 G 0,0 vm 0 v G 0,0, G 1 = P. 19

21 Theorem 9. In the first exceptional case, one has G 0 = G 0,0 PVG 0,1 G 0,1 VP ( G 0,0 PVG 0,1 ) vm 0 v ( G 0,0 G 0,1 VP ) + PVG 0,2 VP, G 1 = P + PVG 0,3 VP + PVG 0,2 VPVG 0,2 VP [ I ( G 0,0 PVG 0,1 ) vm 0 v ]( I + G 0,1 VPV ) G 0,2 VP PVG 0,2 ( I + VPVG0,1 )[ I vm 0 v ( G 0,0 G 0,1 VP )]. Theorem 10. In the second exceptional case, one has G 0 = (I P) ( G 0,0 G 0,0 vm 0 v G 0,0 ) (1 P), G1 = P. 20

22 Let Q B(K) be the orthogonal projections onto Ker M 0, and m 0 = N Qv n (α) Qv n (α). α=1 Theorem 11. In the third exceptional case, one has G 0 = (I P) [ G 0,0 G 0,0 vm 0 v G 0,1 G 0,1 vm 0 v G 0,0 ( ) G 0,0 I vm 0 v G 0,1 vm 0 v ( I G 0,1 vm 0 v ) G 0,0 + G 0,0 vm 0 v G 0,2 vm 0 v G 0,0 + G 0,0 vm 0 v G 0,0 PG 0,0 vm 0 v G 0,0 ]( I P ), G 1 = P + PVG 0,3 VP PVG 0,2 vm 0 v G 0,2 VP (I P) [ I G 0,0 ( I vm 0 v G 0,1 ) vm 0 v ]( I G 0,1 vm 0 v ) G 0,2 VP PVG 0,2 ( I vm 0 v G 0,1 )[ I vm 0 v ( I G 0,1 vm 0 v ) G 0,0 ] (I P). 21

23 Idea for the proofs By the resolvent equation R(κ) = R 0 (κ) R(κ)VR 0 (κ) = R 0 (κ) R 0 (κ)vr(κ) = R 0 (κ) R 0 (κ)v [ U Uv R(κ)vU ] v R 0 (κ). Note that U Uv R(κ)vU = [ U + v R 0 (κ)v ] 1. Hence, if we set M(κ) = U + v R 0 (κ)v, R(κ) = R 0 (κ) R 0 (κ)vm(κ) 1 v R 0 (κ). Thus it suffices to expand M(κ) 1. 22

24 Proposition 12. Assume (A) for an integer β 2. Then, M(κ) = β 2 j=0 where M j B(K) is given by κ j M j + O(κ β 1 ) in B(K), M 0 = U + v G 0,0 v, M j = v G 0,j v for j = 1, 2,.... If M 0 is invertible, we can use the Neumann series to expand M(κ) 1. Otherwise, the essential idea is the following (although it is not exactly what we shall do in our paper): 23

25 Let Q: K Ker M 0 be the orthogonal projection, and consider the block decomposition associated with K = (1 Q)K QK: ( ) ( ) M0 0 X(κ) Y(κ) M(κ) = + κ 0 0 Y(κ) Z(κ) = I 0 κy(κ) [ M 0 + κx(κ) ] 1 I M 0 + κx(κ) 0 0 κz(κ) κ 2 Y(κ) [ M 0 + κx(κ) ] 1 Y(κ) I κ [ M 0 + κx(κ) ] 1 Y(κ) 0 I If the leading term Z 0 of Z(κ) is invertible in B(QK), we can expand M(κ) 1. Otherwise, let S: QK Ker Z 0 be the orthogonal projection, and repeat the arguments on QK = (1 S)QK SQK.. 24

26 Model operator Assumption (B). Assume A(κ) B(K), κ D C \ {0}, satisfy 1. D C \ {0} is invariant under the complex conjugation, and accumulates on 0 C. 2. For each κ D the operator A(κ) satisfies A(κ) = A(κ) and has an inverse A(κ) 1 B(K). 3. As κ 0 in D, in the norm topology of B(K) A(κ) = A 0 + κã1(κ) with Ã1(κ) = O(1). 4. σ(a 0 ) does not accumulate on 0 C as a set. 25

27 Expansion scheme of Jensen Nenciu (2001) Proposition. Assume (B). Let Q B(K) be the orthogonal projection onto Ker A 0, and define a(κ) B(QK) by a(κ) = 1 { Q Q(A(κ) + Q) 1 Q } κ = ( 1) j κ j QÃ1(κ) [ (A 0 + Q)Ã1(κ) ] j Q. j=0 Then a(κ) is bounded in B(QK) as κ 0 in D. Moreover, for κ D close to 0 the operator a(κ) is invertible in B(QK), and A(κ) 1 = (Q + A(κ)) κ (Q + A(κ)) 1 a(κ) (Q + A(κ)) 1. 26